---
tag: 7Q6KK8WJ02
kind: definition
name: Ore condition
msc: [16U20, 16S85]
defines: TJP4ARC2X7
uses: [0B1261B5ED, AN9SKEZ3VY, 0NNRDW10RZ]
sources:
  - url: https://en.wikipedia.org/wiki/Ore_condition
---

## Definition

Let $R$ be a [[ring|0B1261B5ED]] with identity.
Let $S \subseteq R$ be a [[multiplicative set|AN9SKEZ3VY]]: $1 \in S$ and $st \in S$ for all $s, t \in S$.
The set $S$ satisfies the *right Ore condition* if for all $a \in R$ and $s \in S$ there exist $b \in R$ and $t \in S$ with
$$a t = s b.$$
Equivalently, $aS \cap sR \ne \emptyset$ for all $a \in R$ and $s \in S$.
The *left Ore condition* asks instead for $b \in R$ and $t \in S$ with $t a = b s$.
A *right Ore domain* is a domain $R$ such that $R \setminus \set{0}$ satisfies the right Ore condition.

## Remarks

Example: if $R$ is commutative, every multiplicative set $S$ satisfies both Ore conditions: take $t = s$ and $b = a$.
Every right Noetherian domain is a right Ore domain.

Non-example: let $k$ be a field and $R = k\langle x, y \rangle$ the free algebra, with $S = R \setminus \set{0}$.
Each nonzero element of $xS$ is a combination of words beginning with $x$, and each nonzero element of $yR$ is a combination of words beginning with $y$.
So $xS \cap yR = \set{0}$, while $0 \notin xS$ since $R$ is a domain.
Hence $xS \cap yR = \emptyset$ and the right Ore condition fails for $a = x$ and $s = y$.

The right Ore condition says that a left fraction $s^{-1} a$ can be rewritten as a right fraction $b t^{-1}$, since $a t = s b$ gives $s^{-1} a = b t^{-1}$ once $s$ and $t$ are invertible.
A multiplicative set that satisfies the right Ore condition and is right reversible (if $s a = 0$ with $s \in S$, then $a t = 0$ for some $t \in S$) is a *right denominator set*.
If $S$ contains no [[zero divisors|0NNRDW10RZ]], reversibility is automatic.